A 79.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her

Question

A 79.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the cord is 10.0 m. The jumper reaches reaches the bottom of her motion 33.0 m below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 10.0-m free-fall and a 23.0-m section of simple harmonic oscillation.

(a) For the free-fall part, what is the appropriate analysis model to describe her motion.

a)particle under constant angular acceleration

b)particle in simple harmonic motion       or

c)particle under constant acceleration

(b) For what time interval is she in free-fall?
s

(c) For the the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or non-isolated?

A)isolated     or

B)non-isolated   

(d) From your response in part (c) find the spring constant of the bungee cord.
N/m

(e) What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper?
m below the bridge

(f) What is the angular frequency of the oscillation?
rad/s

(g) What time interval is required for the cord to stretch by 23.0 m?
s

(h) What is the total time interval for the entire 33.0-m drop?
s

Explanation / Answer

a) b)particle in simple harmonic motion

b) 10=.5*9.81*t^2
solve for t
1.4278 s

c) A)isolated

d)
79*9.81*33=.5*k*23^2
solve for k
k=2*79*9.81*33/23^2
96.69 N/m

e)
96.69*x=79*9.81
x=79*9.81/96.69
x=8.0152
this is 33-8.0152 m above the max stretch
f)
?=sqrt(k/m)
1.1063 rad/s
g)
T=2*?/?
5.6765 s

h) 1.4278+5.6765
7.1043 s

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